pub fn new_birthday_probability(n: u32) -> f64 {
    //计算n个人中至少两人生日相同的概率,用365处理
    let mut p = 1.0;
    for i in 0..n {
        p *= (365 - i) as f64 / 365.0;
    }
    return 1.0 - p
}
